Re: Finest partition - exercise in Suppes's book


MoeBlee wrote:
> smnewberger@xxxxxxxxxxx wrote:
> > Your next to final sentence starting "Then" shows that every member
> > of P is a singleton.
> > Your last sentence is a little unclear -I do not know what Q is
>
> Oops. Typo. Should have been P, not Q. Here's the last bit with
> correction:
>
> Suppose:
> P not= ({y}| yeS} &
> P is a partition of S &
> AQ((Q not= P & Q is a partition of S) -> P is finer per S than Q).
>
> Then Ax(xeP -> Eb(be{y}| yeS} & x subsetof b).
>
> If P not= {{y}| yeS}, then there is some non-empty x that's a member of
> P but x is not a singleton. So x cannot be a subset of a singleton.
>
> Look okay now?
Yes smn
> > This means with "finer"just a general relation- 1) if P is
> > finer than Q then Q is not finer then P and 2) if P is finer then Q and
> > Q finer then R then P is finer then R. ( 3) P is not finer then P
>
> Right, It's irreflexive, asymmetric, and transitive.
>
> Thanks for looking at my original post.
>
> MoeBlee

.

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